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Question
A real value of x satisfies the equation \[\frac{3 - 4ix}{3 + 4ix} = a - ib (a, b \in \mathbb{R}), if a^2 + b^2 =\]
Options
1
-1
2
-2
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Solution
\[a - ib = \frac{3 - 4ix}{3 + 4ix}\]
\[ = \frac{3 - 4ix}{3 + 4ix} \times \frac{3 - 4ix}{3 - 4ix}\]
\[ = \frac{9 + 16 x^2 i^2 - 24xi}{9 - 16 x^2 i^2}\]
\[ = \frac{\left( 9 - 16 x^2 \right) - i\left( 24x \right)}{9 + 16 x^2}\]
\[ \Rightarrow \left| a - ib \right|^2 = \left| \frac{\left( 9 - 16 x^2 \right) - i\left( 24x \right)}{9 + 16 x^2} \right|^2 \]
\[ \Rightarrow a^2 + b^2 = \frac{\left( 9 - 16 x^2 \right)^2 + \left( 24x \right)^2}{\left( 9 + 16 x^2 \right)^2}\]
\[ = \frac{81 + 256 x^4 - 288 x^2 + 576 x^2}{\left( 9 + 16 x^2 \right)^2}\]
\[ = \frac{81 + 256 x^4 + 288 x^2}{\left( 9 + 16 x^2 \right)^2}\]
\[ = \frac{\left( 9 + 16 x^2 \right)^2}{\left( 9 + 16 x^2 \right)^2}\]
\[ = 1\]
Hence, the correct option is (a).
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